One out of two, three out of three 1, one out of four, five out of five 1, and six out of six 1.

So if you subtract one, it can be divisible by 2, 3, 4, 5 and 6.

The least common multiple of 2, 3, 4, 5 and 6 is 60.

So the number is 60a+ 1.

Divisible by 7.

60a+ 1=7n

n =(60a+ 1)/7 = 8a+(4a+ 1)/7

Order (4a+ 1)/7=b

Then 4a+ 1=7b.

a =(7 B- 1)/4 = b+(3 B- 1)/4

Order c=(3b- 1)/4

4c=3b- 1

b =(4c+ 1)/3 = c+(c+ 1)/3

So if c is divided by 3, the remainder is 2.

So c=3m+2.

b=4m+3

a=7m+5

n=(60a+ 1)/7=60m+43

60a+ 1=7n=420m+30 1

So there are countless solutions.

M=0 is the minimum value, with 420×0+30 1=30 1.